Example 55.14.1. Let $R$ be a discrete valuation ring with uniformizer $\pi $. Given $n \geq 0$, consider the ring map
Set $X = \mathop{\mathrm{Spec}}(A)$ and $S = \mathop{\mathrm{Spec}}(R)$. If $n = 0$, then $X \to S$ is smooth. For all $n$ the morphism $X \to S$ is at-worst-nodal of relative dimension $1$ as defined in Algebraic Curves, Section 53.20. If $n = 1$, then $X$ is regular, but if $n > 1$, then $X$ is not regular as $(x, y)$ no longer generate the maximal ideal $\mathfrak m = (\pi , x, y)$. To ameliorate the situation in case $n > 1$ we consider the blowup $b : X' \to X$ of $X$ in $\mathfrak m$. See Divisors, Section 31.32. By construction $X'$ is covered by three affine pieces corresponding to the blowup algebras $A[\frac{\mathfrak m}{\pi }]$, $A[\frac{\mathfrak m}{x}]$, and $A[\frac{\mathfrak m}{y}]$.
The algebra $A[\frac{\mathfrak m}{\pi }]$ has generators $x' = x/\pi $ and $y' = y/\pi $ and $x'y' = \pi ^{n - 2}$. Thus this part of $X'$ is the spectrum of $R[x', y'](x'y' - \pi ^{n - 2})$.
The algebra $A[\frac{\mathfrak m}{x}]$ has generators $x$, $u = \pi /x$ subject to the relation $xu - \pi $. Note that this ring contains $y/x = \pi ^ n/x^2 = u^2\pi ^{n - 2}$. Thus this part of $X'$ is regular.
By symmetry the case of the algebra $A[\frac{\mathfrak m}{y}]$ is the same as the case of $A[\frac{\mathfrak m}{x}]$.
Thus we see that $X' \to S$ is at-worst-nodal of relative dimension $1$ and that $X'$ is regular, except for one point which has an affine open neighbourhood exactly as above but with $n$ replaced by $n - 2$. Using induction on $n$ we conclude that there is a sequence of blowing ups in closed points
such that $X_{\lfloor n/2 \rfloor } \to S$ is at-worst-nodal of relative dimension $1$ and $X_{\lfloor n/2 \rfloor }$ is regular.
Comments (2)
Comment #2444 by Bronson Lim on
Comment #2487 by Johan on
There are also: